Monte Carlo Estimation of PI in Python

pi-monte-carlo.py

import random as r
import math as m

# Number of darts that land inside.
inside = 0
# Total number of darts to throw.
total = 1000

# Iterate for the number of darts.
for i in range(0, total):
  # Generate random x, y in [0, 1].
  x2 = r.random()**2
  y2 = r.random()**2
    # Increment if inside unit circle.
    if m.sqrt(x2 + y2) < 1.0:
      inside += 1

# inside / total = pi / 4
pi = (float(inside) / total) * 4

# It works!
print(pi)

• I would suggest using random.uniform() instead of random.random() since some darts might be landed on the edge of the square, but random.random() only generates a random float uniformly in the semi-open range [0.0, 1.0), while random.uniform() generates a random float uniformly in the range [0.0, 1.0].

• no indentation needed in the if clause.

• no math.sqrt needed (reduce computational cost). if math.sqrt(x2 + y2) < 1.0 then x2+y2 < 1.0, vice versa.

Doesn't this do the same thing without dealing with the square or the x,y coords?

import random
iterations = 1000000
x = 0
for i in range(iterations):
	x+=(1-random.random()**2)**.5
print(4*x/iterations)

Thank you. Here is a translated SNAP! program version...https://snap.berkeley.edu/project?user=nisar&project=monte_carlo_pi_calculation

Thank you! This helped me understand how to run it with x and y.

Here is a slightly faster version.

def square_pi(N):
    # square_pi(100000000) is fast and gives 4 digits. 
    xy = np.random.random((N, 2)) ** 2
    counts = np.sum((xy[:, 0] + xy[:, 1]) <= 1)

    print("pi was approximated at ::",4*counts/N)
    return 4*counts/N